Question: Evaluate the iterated integral. $ \int_{-1}^1 \left( \int_{-2}^2 e^x + e^y \, dy \right) dx =$ Choose 1 answer: Choose 1 answer: (Choice A) A $e^2 - 4e - 4e^{-1} + e^{-2}$ (Choice B) B $2e^2 + 4e - 4e^{-1} - 2e^{-2}$ (Choice C) C $2e^2 - 2e^{-2}$ (Choice D) D $4e^2 + 4e^{-2}$
Explanation: Evaluate the inner integral: $\begin{aligned} \int_{-1}^1 \left( \int_{-2}^2 e^x + e^y \, dy \right) dx &= \int_{-1}^1 \left[ ye^x + e^y \right]_{-2}^2 \, dx \\ \\ &= \int_{-1}^1 2e^x + e^2 - \left( -2e^x + e^{-2} \right) \, dy \\ \\ &= \int_{-1}^1 4e^x + e^2 - e^{-2} \, dy \end{aligned}$ Evaluate the outer integral: $\begin{aligned} \int_{-1}^1 4e^x + e^2 - e^{-2} \, dy &= 4e^x + x \left( e^2 - e^{-2} \right) \bigg|_{-1}^1 \\ \\ &= 4e + e^2 - e^{-2} - \left( 4e^{-1} - e^2 + e^{-2} \right) \\ \\ &= 2e^2 + 4e - 4e^{-1} - 2e^{-2} \end{aligned}$ The answer: $ \int_{-1}^1 \left( \int_{-2}^2 e^x + e^y \, dy \right) dx = 2e^2 + 4e - 4e^{-1} - 2e^{-2}$